HDU Phone List dictionary tree prefix judgment

Chain: http://acm.hdu.edu.cn/showproblem.php?pid=1671

Phone List Time limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 12763 Accepted Submission (s): 4341


Problem Description Given A list of phone numbers, determine if it is consistent in the sense this no number is the prefix of another. Let ' s say the phone catalogue listed these numbers:
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it's not possible-to-call Bob, because-the central would direct your-to-the-emergency line as soon as Y OU had dialled the first three digits of Bob ' s phone number. So the list would not being consistent.

Input the first line of input gives a single integer, 1 <= t <=, the number of test cases. Each test case is starts with N, the number of the phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with a unique phone number on each line. A phone number is a sequence of at the most ten digits.
Output for each test case, output "YES" If the list is consistent, or "NO" otherwise.
Sample Input

2 3 911 97625999 91125426 5 113 12340 123440 12345 98346
Sample Output

NO YES

Test instructions: Gives a string of n strings, asking if there will be a string that is prefixed with another string.

Practice: Constantly add to the trie, and then constantly judge whether the prefix of other strings, or the existing is not their own prefix.

#include <stdio.h> #include <string.h> #include <stdlib.h> #define MAX struct Trie {Trie *ne   
    Xt[max];   int V;   
 
Change as needed};

Trie Root;
    void Createtrie (char *str) {int len = strlen (str);
    Trie *p = &root, *q;
        for (int i=0; i<len; ++i) {int id = str[i]-' 0 ';
            if (p->next[id] = = NULL) {q = (Trie *) malloc (sizeof (Trie));    Q->v = 1;
            Initial v==1 for (int j=0; j<max; ++j) q->next[j] = NULL;
            P->next[id] = q;
        p = p->next[id];
            } else {p->next[id]->v++;
        p = p->next[id];   }} p->v =-1;
    At the end, the V is changed to 1 for the} int Findtrie (char *str) {int len = strlen (str);
	Trie *p = &root;
    if (P==null) return 0;
        for (int i=0; i<len; ++i) {int id = str[i]-' 0 ';
        p = p->next[id];
    if (p = = NULL)//If it is an empty set, the string that is prefixed by this is not saved        return 0;
		if (p->v = = -1&&i==len-1)//have the same string return-1;
	if (p->v = =-1)//The string in the character set is the prefix of this string return-1;   } return-1;
    This string is the prefix of a string in the character set} int Dealtrie (trie* T) {int i;
    if (T==null) return 0;
			for (i=0;i<max;i++) {if (t->next[i]!=null) {Dealtrie (t->next[i]);
		t->next[i]=null;
    }} if (T!=&root) free (T);
return 0;
} Char str[100]; 
	int main () {int t;
	scanf ("%d", &t);
		while (t--) {int n;
		scanf ("%d", &n); 
		int flag=1;
			while (n--) {scanf ("%s", str);
			int Tem=findtrie (str);
			if (tem==-1) flag=0; 
		Createtrie (str);
		} if (flag==0) puts ("NO"); 
		Else puts ("YES");
	Dealtrie (&root);
} return 0; }